a00 x 11=aa00    

b0 x 11 =  bb0 

aa00+ bb0    = 6710 => 0+b =1=> b= 1

aa00+ 110    = 6710 =>  aa00   = 6710 -110=6600 => a=6

Vậy b =1 ; a =6

a=3

b=0

\(x-\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...-\frac{20}{53.55}=\frac{3}{11}\)

Đặt \(A=\frac{20}{11.13}-\frac{20}{13.15}-\frac{20}{15.17}-...+\frac{20}{53.55}\)

\(A=\frac{10.2}{11.13}-\frac{10.2}{13.15}-\frac{10.2}{15.17}-...-\frac{10.2}{53.55}\)

\(A=10\left(\frac{2}{11.13}-\frac{2}{13.15}-\frac{2}{15.17}-...-\frac{2}{53.55}\right)\)

\(A=10\left(\frac{1}{11}-\frac{1}{55}\right)\)

\(A=10.\frac{4}{55}\)

\(A=\frac{40}{55}\)\(A=10.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{53}-\frac{1}{55}\right)\)

Sửa đề: \(D=\frac{1}{10\cdot11}+\frac{1}{11\cdot12}+\cdots+\frac{1}{99\cdot100}\)

\(=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\cdots+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{10}-\frac{1}{100}=\frac{10}{100}-\frac{1}{100}=\frac{9}{100}\)

=>D là số hữu tỉ

\(C=0,3\cdot12,8+0,3\cdot7,2\)

\(=0,3\cdot\left(12,8+7,2\right)\)

\(=0,3\cdot20=6\)

=>C là số hữu tỉ

\(E=\frac{\frac{4}{11}+\frac{4}{121}-\frac{4}{12321}}{\frac{9}{11}+\frac{9}{121}-\frac{9}{12321}}=\frac{4\left(\frac{1}{11}+\frac{1}{121}-\frac{1}{12321}\right)}{9\left(\frac{1}{11}+\frac{1}{121}-\frac{1}{12321}\right)}=\frac49\)

=>E là số hữu tỉ

Ta có: \(B=\frac{4}{15}\cdot\frac{-25}{24}=\frac{4\cdot\left(-25\right)}{15\cdot24}=\frac{-100}{360}=\frac{-5}{18}\)

=>B là số hữu tỉ